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I am trying to prove that $\Bbb{R}^n$ minus finitely many points $x_1,\ldots,x_m$ is simply connected, where $n \geq 3$. For days now I have tried many different arguments but I have found flaws in all of them. I have finally come up with one, except that there is some small detail that I need to know how to prove.

I prove that $\Bbb{R}^n$ minus finitely many points is simply connected by inducting on the number of points that I remove. If I remove one point (the case $m=1$), I get that $\Bbb{R}^n - \{x_1\} \cong S^{n-1} \times \Bbb{R}$ which upon applying $\pi_1$ shows me that $\Bbb{R}^n - \{x_1\}$ is simply connected.

Inductive Hypothesis: Now suppose that $\Bbb{R}^n -\{x_1,\ldots,x_k\}$ is simply connected for all $ k <m$.

Suppose now I have $m$ points $x_1,\ldots,x_m$ lying in $\Bbb{R}^n$. Now if I write each of these points out in coordinates, I know that there is at least one $j$ with $1 \leq j \leq n$ such that the $j-th$ coordinate of all my points are not all the same (otherwise my points are all just one point and there is nothing to prove!).

Because I have only finitely many points, suppose without loss of generality that $x_1$ is the point whose $j-th$ coordinate is the greatest (this does not mean that such a choice is unique, I only want to know if it exists). Say that the $j-th$ coordinate of $x_1$ is $c$. Now I consider the plane

$$\mathbf{x}_j = \{\mathbf{x} \in \Bbb{R}^n : \text{$j$ -th coordinate of $\mathbf{x}$ is equal to $c$} \}.$$

My idea now is to apply the Seifert-Van Kampen Theorem together with the induction hypothesis as follows: I set

$$A = \Bigg\{\mathbf{x} \in \Bbb{R}^n- \{x_1,\ldots,x_m\} : \text{$j$ -th coordinate of $\mathbf{x}$ is greater than $c-\varepsilon$} \Bigg\}$$

where $\varepsilon$ is chosen such that $A$ does not enclose all my points and

$$B = \Bigg\{\mathbf{x} \in \Bbb{R}^n- \{x_1,\ldots,x_m\} : \text{$j$ -th coordinate of $\mathbf{x}$ is less than $c $} \Bigg\}.$$

Here's a picture of what I'm trying to do in the case of $\Bbb{R}^3$: What I'm trying to do in the case of $\Bbb{R}^3$

Then $A$ is open and so is $B$, clearly $A$ and $B$ are path connected and their intersection which is just a "cuboid" being a convex set is path connected as well.

My Problem: I want to apply my inductive hypothesis to $A$ and $B$ in order to deduce that $\pi_1(A) = \pi_1(B) = 0$. The problem now is that the inductive hypothesis is for $\Bbb{R}^n$ and not "chopped off bits" of $\Bbb{R}^n$ like $A$ and $B$. How do I get around this? Can I say that $A$ and $B$ are somehow deformation retracts of $\Bbb{R}^n$ minus finitely many points?


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Your first claim about distinct coordinates is false. Consider the $2^n$ vertices of the unit cube in $\mathbb{R}^n$. Why not try a simpler way of isolating the punctures? – Zhen Lin Aug 7 '12 at 7:25
I'm not sure if I understand your picture correctly, but if you draw a separating hyperplane between a few points and the others, the resulting two half-spaces are homeomorphic to $\mathbb{R}^{n}$ minus a bunch of points and you're in position to apply induction and Seifert-van Kampen. – t.b. Aug 7 '12 at 7:25
@ZhenLin I don't understand what's the problem. Even if I look at the unit cube in $\Bbb{R}^3$, I can take the plane $z =1$. – user38268 Aug 7 '12 at 7:27
yes, sure, but that's no problem, is it? You have $H = \{x \,:\,\langle n,x \rangle = c\}$ where $n$ is the normal vector, so take $A= \{x \,:\,\langle n,x \rangle \lt c + \varepsilon\}$ and $B = \{x \,:\,\langle n,x \rangle \gt c - \varepsilon\}$ with $\varepsilon$ small enough ($\varepsilon$ should only be smaller than the minimal distance of your points to the hyperplane). – t.b. Aug 7 '12 at 7:33
@BenjaLim: You can still use t.b.'s method to prove $A$ and $B$ in your problem is actually homeomorphic to $\mathbb{R}^n$ minus finitely many points. To see this, all you need is to settle a homeomorphism between any open half space and $\mathbb{R}^n$, then minus those points. – Yuchen Liu Aug 7 '12 at 7:37

2 Answers 2

up vote 3 down vote accepted

The problem is, in fact, a simple induction. The base case with $m = 1$ is easily dealt with, as you did. Now assume that $m > 1$ and divide the points in $S = \{x_1, \ldots, x_m\}$ in two sets of smaller size (no matter how), say $A$ and $B$.

For convenience, let's assume that $A$ and $B$ are separated by the hyperplane $\mathcal{H}$, and that $N_{+}$ and $N_{-}$ are two open neighborhoods of the half-spaces that result. For an arbitrary base-point $x_0 \in \mathcal{H}$, Van Kampen theorem applies, giving a surjection from $\pi_1(N_{+} \backslash A) \ast \pi_1(N_{-} \backslash B)$ to $\pi_1(\mathbb{R}^n - S)$. Now just use the induction hypothesis to conclude that $\pi_1(N_{+} \backslash A) = \pi_1(N_{-} \backslash B) = \pi_1(\mathbb{R}^n - S) = 0$, as you wish.

Edit: Of course that $N_+ \backslash A$ and the other set are homeomorphic (or, if you want, homotopy equivalent) to $\mathbb{R^n} \backslash A$ in an obvious way. Such geometric observations are not a difficulty once you understood how to apply the theorem.

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That's exactly what I am trying to do, just written out in more detail. – user38268 Aug 7 '12 at 7:34
Please see my comment here:… – user38268 Aug 7 '12 at 7:36

We may compactify $\mathbb{R}^{n}$ to be $\mathbb{S}^{n}$ by adding a point, and for $n\ge 3$ it should not influence the statement. Now we can proceed inductively since $S^{n}$ removed $k$ points should be homeomorphic to $\bigvee^{k-1}_{i=1} S^{n-1}_{i}$. For $n\ge 3$ we are left with spheres of dimension 2 or higher, so we can conclude the fundamental group must be trivial. It should not be too difficult to visualize.

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